Justification for $f_ig_j \equiv 0$ in $U \cup V = \mathbb C^n$ for varieties $U$ and $V$.

Question

Artin Algebra Chapter 11

Here are solutions by Takumi Murayama here and Brian Bi. Here is the definition of a variety in $\mathbb C^n$

Here is Takumi Murayama's solution for $U \cup V = \mathbb C^n$

1. I have questions about the finite zeroes part.

Why are there finite zeroes? Is this a generalization of the following theorem for 2 polynomials and $\mathbb C^2$ to finitely many polynomials and $\mathbb C^n$

?

2. I have a question about a different argument.

Can we instead argue that the only such polynomial in the variety $\mathbb C^n$ is the zero polynomial?

My argument is that $\mathbb C^n$ as a variety is defined by $$\mathbb C^n := \{\text{the zero polynomial}=\text{the number zero}\}$$

So without references to finiteness, we must have $f_ig_j \equiv \text{the zero polynomial} \ \forall i,j$. Then we conclude $U=\mathbb C^n$ or $V = \mathbb C^n$.

Answer 1

Murayama's solution is not correct as stated: in $\Bbb C^2$, the polynomial $x$ has the $y$-axis as it's zeroes, which is an infinite set. A statement which is close to Murayama's which is correct would be that the zeroes of a nonconstant polynomial consist of a finite number of irreducible components. These irreducible components are small, and it cannot be the case that a union of finitely many of them is the whole of $\Bbb C^n$ - there are many different proofs (they're meager and one may apply the Baire category theorem, there are proofs from the perspective of basic ring theory, dimension-theoretic proofs, etc).

For your second question, there are some issues with the language, but the idea is essentially correct. It would be best to say that "the only set* of polynomials which defines $\Bbb C^n$ is the set consisting of the zero polynomial". This statement is correct, and the justification is that every nonconstant polynomial over $\Bbb C$ must be nonvanishing at some point (for example, via the fundamental theorem of algebra), and therefore cannot be in the set defining $\Bbb C^n$.

(*) soon you will probably be more used to speaking about ideals defining varieties than sets of polynomials, and the correct adjustment to the language is that the ideal of $\Bbb C^n$ is exactly the zero ideal.